I am trying to prove the following and want to see if I am on the right track.
Let $V$ be a finite-dimensional inner product space with dimension $n$ over either the field $\mathbb{F} = \mathbb{C}$ or $\mathbb{F} = \mathbb{R}$. Then there exists an isometry $T : V \rightarrow \mathbb{F}^n$, where $\mathbb{F}^n$ is equipped with the standard inner product.
My idea is the following. Let $v_1, \dotsc, v_n$ be an orthonormal basis of $V$. Write some $v \in V$ as $v = a_1v_1 + \cdots + a_nv_n$. Then $\Vert v \Vert^2 = \langle v, v \rangle = \lvert a_1 \rvert^2 + \cdots + \lvert a_n\rvert^2$.
It seems the desired isometry $T$ should be defined by $v \mapsto \lvert a_1 \rvert e_1 + \cdots + \lvert a_n \rvert e_n$, where $e_j$ is the $j$th standard basis vector in $\mathbb{F}^n.$ Then we would indeed have $\langle Tv, Tv \rangle = \langle v, v\rangle.$ But I do not think this works, since $T$ is not injective.
How can I fix this argument? Or is there a better, different way to approach this problem?